为什么不能在带有模板的派生类中使用基类的别名?

本文介绍了为什么不能在带有模板的派生类中使用基类的别名?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

考虑以下C ++代码:

Consider this C++ code :

template<typename Session> class Step { public: using Session_ptr = boost::shared_ptr<Session>; protected: Session_ptr m_session; public: inline Step(Session_ptr session) : m_session(session) {} }; template<typename Socket> class Session { public: Socket a; Session(Socket _a): a(_a) {} }; template <typename Socket> class StartSession : public Step<Session<Socket> > { protected: Session_ptr m_session; //Unknown type Session_ptr public: inline StartSession(Session_ptr session) : Step<Session<Socket> >(session) {} void operator()(const boost::system::error_code& ec); }; template <typename Socket> class StartSession2 : public Step<Session<Socket> > { protected: typename Step<Session<Socket> >::Session_ptr m_session; public: inline StartSession2(typename Step<Session<Socket> >::Session_ptr session) : Step<Session<Socket> >(session) {} void operator()(const boost::system::error_code& ec); }; int main(int argc, char * argv[]) { Step<Session<int> >::Session_ptr b(new Session<int>(5)); //no problem StartSession<int >::Session_ptr bb(new Session<int>(5)); //gcc ok, clang refuses to remember the symbol since the class has errors StartSession2<int >::Session_ptr bbb(new Session<int>(5)); //no problem std::cout << b->a; // ok std::cout << bb->a; // gcc ok, clang bb not declared std::cout << bbb->a; // ok return 0; }

如您所见，这里发生了一些奇怪的事情(至少对我来说)...

As you can see, there are some strange (to me at least) things happening here...

首先，为什么在子类中不能访问Session_ptr? 我知道，因为这些是模板化的类，这会使事情变得更复杂...但是我看不出有任何强制使用typename的歧义...

First, why isn't Session_ptr accessible in the child classes ? I know because these are templated class, that make things more complicated... But I don't see any ambiguity here that makes the use of typename mandatory...

然后，为什么在主目录中Session_ptr既可以作为基类的成员又可以作为子类的成员来访问?

Then, why in the main, Session_ptr is accessible either as member of the base class either as member of a child class ?

推荐答案

不合格的查找不在类模板的依赖基类中查找.

Unqualified lookup does not look in dependent base classes in class templates.

所以在这里:

template <typename Socket> class StartSession : public Step<Session<Socket> > { protected: Session_ptr m_session; // <== unqualified name lookup on Session_ptr // ... };

Step<Session<Socket>>是StartSession<Socket>的依赖基类.为了在那里查找，您必须进行 qualified 名称查找(这是您在StartSession2中所做的事情):

Step<Session<Socket>> is a dependent base class of StartSession<Socket>. In order to lookup there, you'll have to do qualified name lookup (which is what you're doing in StartSession2):

template <typename Socket> class StartSession : public Step<Session<Socket> > { protected: typename Step<Session<Socket>>::Session_ptr m_session; // ... };

或者只是自己添加别名:

Or simply add the alias yourself:

using Session_ptr = typename Step<Session<Socket>>::Session_ptr;